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# Number of ways to choose an integer such that there are exactly K elements greater than it in the given array

• Difficulty Level : Easy
• Last Updated : 28 May, 2021

Given an array arr[] of N elements and an integer K, the task is to find the number of ways of choosing an integer X such that there are exactly K elements in the array that are greater than X.
Examples:

Input: arr[] = {1, 3, 4, 6, 8}, K = 2
Output:
X can be chosen as 4 or 5
Input: arr[] = {1, 1, 1}, K = 2
Output:
No matter what integer you choose as X, it’ll never have exactly 2 elements greater than it in the given array.

Approach:
We count total number of distinct elements in the array. If the count of distinct elements is less than or equal to k, then 0 permutations possible. Else count is equal to number of distinct elements minus k.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to returns the required count of integers``int` `countWays(``int` `n, ``int` `arr[], ``int` `k)``{` `    ``if` `(k <= 0 || k >= n)``        ``return` `0;` `    ``unordered_set<``int``> s(arr, arr+n);``    ``if` `(s.size() <= k)``       ``return` `0;` `    ``// Return the required count``    ``return` `s.size() - k;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 100, 200, 400, 50 };``    ``int` `k = 3;``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << countWays(n, arr, k);``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to returns the``// required count of integers``static` `int` `countWays(``int` `n, ``int` `arr[], ``int` `k)``{` `    ``if` `(k <= ``0` `|| k >= n)``        ``return` `0``;``    ``Set s = ``new` `HashSet();``    ``for``(``int` `i = ``0``; i < n; i++)``        ``s.add(arr[i]);``        ` `    ``if` `(s.size() <= k)``        ``return` `0``;` `    ``// Return the required count``    ``return` `s.size() - k;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `arr[] = { ``100``, ``200``, ``400``, ``50` `};``    ``int` `k = ``3``;``    ``int` `n = arr.length;``    ``System.out.println(countWays(n, arr, k));``}``}` `// This code id contributed by``// Surendra_Gangwar`

## Python3

 `# Python 3 implementation of the approach` `# Function to returns the required count of integers``def` `countWays(n, arr, k) :` `    ``if` `(k <``=` `0` `or` `k >``=` `n) :``        ``return` `0` `    ``s ``=` `set``()``    ``for` `element ``in` `arr :``        ``s.add(element)``        ` `    ``if` `(``len``(s) <``=` `k) :``        ``return` `0``;` `    ``# Return the required count``    ``return` `len``(s) ``-` `k;`  `# Driver code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``arr ``=` `[ ``100``, ``200``, ``400``, ``50` `]``    ``k ``=` `3``;``    ``n ``=` `len``(arr) ;``    ``print``(countWays(n, arr, k))` `# This code is contributed by Ryuga`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `    ``// Function to returns the``    ``// required count of integers``    ``static` `int` `countWays(``int` `n, ``int` `[]arr, ``int` `k)``    ``{` `        ``if` `(k <= 0 || k >= n)``            ``return` `0;``        ``HashSet<``int``> s = ``new` `HashSet<``int``>();``        ``for``(``int` `i = 0; i < n; i++)``            ``s.Add(arr[i]);` `        ``if` `(s.Count <= k)``            ``return` `0;` `        ``// Return the required count``        ``return` `s.Count - k;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String []args)``    ``{``        ``int` `[]arr = { 100, 200, 400, 50 };``        ``int` `k = 3;``        ``int` `n = arr.Length;``        ``Console.WriteLine(countWays(n, arr, k));``    ``}``}` `// This code is contributed by Rajput-Ji`

## PHP

 `= ``\$n``)``        ``return` `0;` `    ``\$s` `= ``array``();``    ``foreach` `(``\$arr` `as` `\$value``)``        ``array_push``(``\$s``, ``\$value``);``    ``\$s` `= ``array_unique``(``\$s``);``        ` `    ``if` `(``count``(``\$s``) <= ``\$k``)``        ``return` `0;` `    ``// Return the required count``    ``return` `count``(``\$s``) - ``\$k``;``}` `// Driver code``\$arr` `= ``array``(100, 200, 400, 50);``\$k` `= 3;``\$n` `= ``count``(``\$arr``);``print``(countWays(``\$n``, ``\$arr``, ``\$k``));` `// This code is contributed by mits``?>`

## Javascript

 ``
Output:

`1`

Time Complexity: O(N * log(N))

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